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Luden [163]
3 years ago
9

Please help me answer this thank you

Mathematics
2 answers:
nexus9112 [7]3 years ago
5 0
Scale for y - 1:1
Scale for x - 1:10

Reason- so that i can plot the points on the graph without any modification to the given one.

Now plot the points on the graph using the scale. I cant obviously do that here.
max2010maxim [7]3 years ago
4 0
Scale for y - 1:1
Scale for x - 1:10

Reason- so that i can plot the points on the graph without any modification to the given one.

Now plot the points on the graph using the scale. I cant obviously do that here.1
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Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th
jonny [76]

Answer:

The answer is \sqrt{\frac{6}{5}}

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy

Solving:

\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy

[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.

Replacing the limits:

6*\frac{1}{3} =2

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ \frac{1}{m}

Solving the double integral with these new limits we have:

\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy

This part is a little bit tricky so let's solve the integral first for dy:

\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx

Replacing the limits:

\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx

Solving now for dx:

[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.

Replacing the limits:

\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}

As I mentioned before, this volume is equal to 1, hence:

\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }

3 0
2 years ago
A number is 6 less than its square. find all such numbers?
cestrela7 [59]
Hello There!

2² - 6 = -2
3² - 6 = 3
4² - 6 = 10

Etc....the list goes on

Hope This Helps You!
Good Luck :) 

- Hannah ❤
3 0
3 years ago
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X<br> x2 = 36<br> What is the positive solution
Pavlova-9 [17]
If you mean x^2=36, then the answer is -6 or 6
8 0
3 years ago
PLEASE HELP ASAP PLEEAASSEEE
STatiana [176]
I think It’s A took the test
8 0
3 years ago
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An equation that represents the path of a diver jumping off a diving board is y = –7x2 + 5x + 16. Graph this function from the p
ANTONII [103]

Answer:

See explanation

Step-by-step explanation:

An equation that represents the path of a diver jumping off a diving board is

y = -7x^2 + 5x + 16.

The diver jumped to the water when x = 0.

Then

y=-7\cdot 0^2+5\cdot 0+16=16

The jumper reached the water when y = 0, then

-7x^2+5x+16=0\\ \\D=5^2-4\cdot (-7)\cdot 16=25+448=473\\ \\x_{1,2}=\dfrac{-5\pm\sqrt{473}}{-14}

So, point where the  jumper reached the water is

\left(\dfrac{5+\sqrt{473}}{14},0\right)\approx (1.93,0)

7 0
3 years ago
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